Simplify log_10(11) * log_11(12) * ... * log_999(1000)

[axrw]

I don't know why I can't get these.

I am supposed to simplify this:

\(\displaystyle \log_{10} 11 \ \cdot\ \log_{11} 12\ \cdot\ \cdot\ \cdot\ \log_{998} 999\ \cdot\ \log_{999} 1000\)

According to the back of the book it's 3, and I'm sure there is some ingenious method of getting there without actually doing all of that, but I can't see it.

 

[arthur ohlsten]

let us look at first 3 terms
I will use log for log to base 10

1)
log[11]

2 )
logbase 11[12]=a raising to power of 11
12=11^a take log of each side
log12 =a log 11 solve for a
a = log12/log11
logbase11[12]=log12/log11

3)
logbase12[13]=b raise to power of 12
13= 12^b take log of each side
log13=blog12
b=log13/log12 or
logbase12[13]=log13/log12

multiplying first 3 terms
log11*[log12/log11]*[log13/log12]
log13 answer for first 3

4)
log11*logbase11[12]*logbase12[13]....logbase998[999]logbase999[1000]
log11*[log12/log11]*[log13/log12]...[log999/log998]*[log1000/log999]
simplify by cancellation
log1000=3 answer

Arthur

 

[axrw]

Thank you!

That was really bugging me. I knew I was missing something, but I just couldn't imagine how you would go about it.

 

[arthur ohlsten]

I suggest you look for simple solutions, as you implied,"must be a simpler way".
I guess you are studying how to convert logs from one base to another.
THE INSTRUCTOR ALWAYS {I BELIEVE} gives simple problems to practice the subject being taught.

Good luck in your studies
Arthur

 

[axrw]

In this case there isn't an instructor. I'm just working through a book on my own. I've always been horrible at math and decided I wanted to improve. I did work through all of the simpler problems, and had no problem with them, but upon arriving at that one it just wasn't coming to me. Like I a said, I'm just not mathematically inclined. :/

Thanks.

 

[arthur ohlsten]

Glad your studying math on your own.
I assume the section that the problem was in reference to , was changing bases of logs. The one always used is translating between base e , the natural logs , and base 10. That is going from Ln to log. The problem you had was going from a base log to Log base 10.
Glad I was of some help.
Arthur